Topology and geometry

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I was asked to prove the weak and the strong topology on is different when , in the sense that a sequence converges weakly to on probably not have its norm converge to as well. The hint is the proof on does not work at here.

Conway suggested that for most Banach spaces, is almost never separable unless is reflexive and separable in the first place. So since we know spaces (or in general, spaces over -finite measure space) should be separable, we reach a tautology on the weak star topology on . Since we already know spaces are reflexive, this does not tell us anything as the weak star topology coincide with the strong topology by considering the double dual.

The problem at here, though, is weak topology itself is not the same as strong topology on . So we had two strategies:

1) To construct a sequence such that , but nevertheless satisfies for all .

2) Find a gap in Conway.

1) can be done explicitly in the case by considering sequences whose th term be a constant and the rest be . In this case we know the sequence’s norm is , while it is obviously weakly convergent to $0$. But I am wondering a deeper reason why this fails. So here is a discussion on 2).

Conway’s proof on is complicated, but can be roughly summed up in 3 steps. First he constructed a family of sets such that is the whole space. Second he managed to construct an equivalent metric on by . This strange twisted metric can be proved to be equivalent to the metric induced by weak star topology (namely, induced by the action of elements in ) without much effort. Now since we assumed the original sequence is weakly convergent in $ in the first place, we can emulate the classical type proof by producing a suitably bad enough element in . This element in effect “chop up” the irrelevant finite part and give a bound on the infinite part. So by manipulating all the 3 parts together he was able to bound the norm of the original sequence altogether.

The essential part of the proof is the construct of the metric. The first part used Baire’s Category Theorem, which carries over in any complete metric space. The third part is non-trivial in the case, since then we do not have an obvious bound on the finite part, but since we know we can manipulate it in some ways by fixing the finite order part. However, it is not clear to me how can we construct a similar metric to make the proof work (even though we know this is a wrong direction). The curious question to me is:

Since we know such a metric does not exist, what kind of metric on can be equivalent to the weak star metric? And does this carries over to case as well when the measure is -finite?